Question

Aflati formula moleculara a unei hidrocarburii ce contine 82,75% C stiind ca 0,93g hidrocarbura formeaza 500 ml vapori la 87 grade C si 720 mm Hg.

Answer 1

PV = niuRT, P = 720mmHg = 720/760 = 0,947 atm

niu = m / miu

=> PVmiu = mRT

=> miu = mRT/PV = 0,93x0,082x(273+87)/0,947x0,5

= 58 g/mol

100% hidrocarbura ..... 82,75% C.. si diferenta.. 17,25% H

58 g/mol ..................... a g C .................................... b g H

                                = 48 g C                           = 10 g H

C = 48/12 = 4

H = 10/1 = 10

=> F.m. = C4H10 , butan